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I found a interesting problem that says: Let $(X,S)$ be a measurable space where $X=mathbb R$ and $S$ is the countable co-countable $sigma$-algebra in $mathbb R$, i.e. $S={Asubsetmathbb R: A lor mathbb R-A is finite or countable}$. So the point is to describe the S-measurable functions $ f:Xto Bbb R^*$, where $Bbb R^*$ are the extended real numbers.

Let $I$ denote the image of $f$. If $I$ is not countable then some $rinmathbb R$ will exist such that $Icap(-infty,r)$ and $Icap[r,infty)$ are both not countable. Then the disjoint preimages of these sets cannot be countable, hence $f$ cannot be meausurable.

We conclude that $I$ must be countable. Then for $xin I$ fibres $f^{-1}({x})$ form a countable partition of $X=mathbb R$, so at least one of these fibres is not countable. If for distinct $x,yin I$ the fibres $f^{-1}({x})$ and $f^{-1}({y})$ are both uncountable then we can choose some $r$ in between $x$ and $y$ such that the disjoint preimages of $(-infty,r)$ and $(r,infty)$ are both uncountable, and again $f$ cannot be measurable.

We conclude that there is exactly one $xin I$ with an uncountable fibre. The complement of this fibre is covered by other fibers. Each of them is a countable set and also the number of these fibers is countable. Conclusion: the complement of the mentioned uncountable fiber is countable, wich means that the uncountable fiber is cocountable.

Our final conclusion: $$ftext{ is measurable if and only if }ftext{ is constant on a cocountable set}$$ In that case the preimages of $(-infty,r)$ with $r$ ranging over $mathbb R$ will all be cocountable or countable.

edit (concerning question of @dan in a comment on this answer)

Let it be that $I$ is an uncountable subset of $mathbb R$.

Let $A:=left{ xinmathbb{R}midleft(-infty,xright)cap Itext{ is countable}right} $
and let $B:=left{ xinmathbb{R}midleft(x,inftyright)cap Itext{ is countable}right} $.

Note that the fact that $I$ is uncountable implies that $Acap B=varnothing$.

It is our aim to prove that $Acup Bneqmathbb{R}$.

So we assume that $Acup B=mathbb{R}$ and from here it is enough
to find a contradiction.

At first hand for the shape of $A$ we see three possibilities: $A=varnothing$,
$A=left(-infty,sright]$ for some $sinmathbb{R}$ or $A=mathbb{R}$.

But if $A=mathbb{R}$ then $I=bigcup_{n=1}^{infty}left(Icapleft(-infty,nright)right)$
is countable as well, so the third possibility falls off.

Then similarly for the shape of $B$ we find two possibilities: $B=varnothing$
or $B=left[i,inftyright)$ for some $iinmathbb{R}$.

Then based on $Acup B=mathbb{R}$ we find that also the possibilities
$A=varnothing$ and $B=varnothing$ fall off.

So $mathbb{R}=Acup B=left(-infty,sright]cupleft[i,inftyright)$
implying that $ileq s$.

But then $Acap B=left[i,sright]neqvarnothing$ and a contradiction
is found.